Obtain the expression for the electric field due to an electric dipole at a point on its axial line.

(N/A) Consider an electric dipole consisting of charges $+q$ and $-q$ separated by a distance $2a$. Let $P$ be a point on the axial line at a distance $r$ from the center $O$ of the dipole.
The electric field at point $P$ due to the charge $+q$ is:
$\overrightarrow{E}_{+q} = \frac{1}{4 \pi \varepsilon_{0}} \frac{q}{(r-a)^{2}} \hat{p}$
The electric field at point $P$ due to the charge $-q$ is:
$\overrightarrow{E}_{-q} = -\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{(r+a)^{2}} \hat{p}$
The total electric field at point $P$ is the vector sum of these fields:
$\overrightarrow{E} = \overrightarrow{E}_{+q} + \overrightarrow{E}_{-q} = \frac{q}{4 \pi \varepsilon_{0}} \left[ \frac{1}{(r-a)^{2}} - \frac{1}{(r+a)^{2}} \right] \hat{p}$
Simplifying the expression:
$\overrightarrow{E} = \frac{q}{4 \pi \varepsilon_{0}} \left[ \frac{(r+a)^{2} - (r-a)^{2}}{(r^{2}-a^{2})^{2}} \right] \hat{p} = \frac{q}{4 \pi \varepsilon_{0}} \left[ \frac{4ar}{(r^{2}-a^{2})^{2}} \right] \hat{p}$
Since the dipole moment $p = q(2a)$,we can write:
$\overrightarrow{E} = \frac{1}{4 \pi \varepsilon_{0}} \frac{2pr}{(r^{2}-a^{2})^{2}} \hat{p}$
For a short dipole where $r >> a$,we have $r^{2}-a^{2} \approx r^{2}$:
$\overrightarrow{E} \approx \frac{1}{4 \pi \varepsilon_{0}} \frac{2pr}{r^{4}} \hat{p} = \frac{2p}{4 \pi \varepsilon_{0} r^{3}} \hat{p}$